plz help with trig?

Question

1. Find the acute angle of A. to the nearest second, when :

a. log cos A.= 9.12575
b. log sin A = 9.91655
c.log cot A= 0.11975
d.log tan A= 0.06323.


2. Simplify the following

sin (90° + x) sin (180° + X) + cos ( 90° + X) cos (180° - X)

Answer 1

1. To solve a through d, you must first change them to exponential form. Remember that log[base b](a) = c converts to b^c = a. You would then calculate the cos inverse (cos^(-1) on your calculator), sin^(-1), and tan^(-1).

For log(cotA) = 0.11975, you would first convert it to exponential form, which would be cotA = 10^(0.11975), and then take the reciprocal of both sides to obtain

1/(cotA) = 1/[ 10^(0.11975) ]

But 1/(cotA) = 1/(cosA/sinA) = sinA/cosA = tanA, so

tanA = 1/[ 10^(0.11975)], at this point you would use tan^(-1)(A).
Remember to have your calculator set to DEG (degree) mode!

2. To simplify this, we will use some or all of these trig identities:

sin(a + b) = sinacosb + sinbcosa
sin(a - b) = sinacosb - sinbcosa
cos(a + b) = cosacosb - sinasinb
cos(a - b) = cosacosb + sinasinb

sin(90 + x) sin (180 + x) + cos(90 + x) cos(180 - x)

[sin(90)cosx + sinxcos(90)] [sin(180)cosx + sinxcos(180)] +
[cos90cosx - sin90sinx] [cos180cosx + sin180sinx]

Note that sin(90) = 1, cos(90) = 0, sin(180) = 0, cos(180) = -1

[cosx + 0] [0 + sinx(-1)] + [0 - sinx] [(-1)cosx + 0]
[cosx] [-sinx] + [-sinx] [-cosx]
-sinxcosx + sinxcosx
0

Your final answer should be 0.

Answer 2

1. the way you'd do these is first you take the antilog by raising 10 to the power of the log. For example, if log x = 2 then x = 10^2 = 100. Then you'd find the inverse trig funtion corresponding to that number. Finally you'd convert it to degrees minutes and seconds. Your calculator can do all that if you know how to use it. Here we go...

a) log cos A = 9.12575
cos A = 1335826332
no solution, because cosines are always between -1 and 1

b) no solution, same reason as a

c) log cot A = .11975
cot A ~=1.317...
A = arccot 1.317 ... ~= 31.19902 degrees ~ = 31 degrees 11 min 56 sec

d) log tan A = ..06323
tan A ~= 1.1567...
A = arctan 1.1567...~=49degrees 9 min 22 sec

2. To do this one you need to remember how the trig fuctions change when the angle is transposed into a different quadrant.
sin (90 + x ) = cos x
sin (180 + x) = -sin x
cos (90 + x) = -sin x
cos (180 - x) = -cos x
Substituting,
cos x (-sin x) + (-sin x)(-cos x) =
-sin x cos x + sin x cos x = 0

Answer 3

1) if(log(cosA)) = 9.125 then cosA = 10^9.125, which is > 1, which is impossible.
b) same thing - no solution
c) cotA = 10^(.11975), or tanA = 10^(-.11975) I get 37.1992 degrees, or 37 degrees 11 minutes 56 seconds
d. tanA = 10^(.06323) = 1.156725. Arctan(1.156725) = 49 degrees 9 minutes 23 seconds

Answer 4

Sin(one hundred eighty-2@)=sin(ninety+@) the place @ skill theata sin2@=cos@ (simply by fact Sin(one hundred eighty-2@) lies in 2nd quadrant so beneficial and sin(ninety+@) additionally lies in 2nd quadrant so it is likewise beneficial) 2sin@cos@=cos@ ( sin2@=2sin@cos@) 2sin@=a million sin@=a million/2 @=pie/6 or pie/6+2n(pie) witch is typical cost